application
The first effects pedal designed specifically to work together with the pickups in a guitar to achieve optimum gain and resonance for all tonal applications. Works with all musical styles.
description
By turning the gain control knob you can add pure, flat-EQ gain, ranging from 6dB to 25dB! The Resonance Switch drops the resonant peak of a single coil pickup up to 5KHz, making a single coil sound like a full and fat humbucker. Features fully discrete Class A circuitry along with true bypass. Comes in a heavy-duty steel chassis and runs on a single 9 volt battery.
of note
For Resonance Switch to work properly, the SFX-01 must be first in the signal chain using passive pickups and a standard guitar cable -- not a wireless unit.
This might end up replacing my homemade FlowerBooster. Thanks Miro
ReplyDeletelooks like a really good booster. I like the resonance switch.
ReplyDeleteIt is nice. Although the idea of 6db minimum is somewhat off-putting.
ReplyDeleteWell. There is always a possibility of adding a classic master volume control to the output :)
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mirosol, where they plugged the pot?? And what pot?
Delete100k log pot - output from the board to lug 3, lug 1 to ground and lug is the new output. :) I've built a few SHOs with added master volume, and it's really good like that.
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lug 2 is new output??
DeleteLug 2 of the master volume 100k log pot to output.
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What would be a good low noise PNP? Would a 2N3906 do?
ReplyDeleteNo C25K pots avilable?
ReplyDeleteI just found this dealer wich ships to Europe:
http://www.ebay.es/itm/Mono-Mixer-Potentiometer-16mm-Reverse-Log-C-Pot-/120746208325?pt=UK_BOI_Electrical_Components_Supplies_ET&var=&hash=item61ce930d5d
You may better get the correct ones and do not change the total charge of the circuit, but if there's no chance try this old trick:
""For 25K RevLog you need 5x25K=125K Lin Pot and 25Kx1.25=31.25K resistor soldered on lugs 2 and 3""
also do not find this potentiometer
ReplyDeletewhat is the solution??
Use linear. That'll work just fine.
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but put a 125k, with 31.25K resistor soldered on lugs 2 and 3 as Bibi said??
ReplyDeleteor a 25k lin?? is the same as the rev log?? I do not think it will make the same effect ...
I really want to build this pedal, but I need to know what potentiometer using ...
ReplyDeleteLinear is not the same as reverse logarithmic - but it will work just as well. I built mine with 20K linear, and it works.
DeleteOnly thing that the taper (linear, logarithmic, reverse logaritmic) affects is the feel of the sweep. Middle position:
Linear: 50%
Log: 10%
Rev. log: 90%
So by using a linear taper pot instead of rev log - you would just have to turn the pot to 2/3 instead of middle position on rev log. That doesn't affect the sound at all. Just how much boost you have on specific part of the sweep.
+m
then what is the advantage of a log Rev?
Deletefrom the middle to end so increases 10%
the linear and thus is more certain
the middle, is that 50% of the gain.
Like i said, it only affects the feel of the sweep. For example; Log in this case would not feel very good, as most of the control would be in the last 10% of the sweep.
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it is!!!
Deleteis better to use 20k lin or 25k lin ???
Whatever you like, but 25K would be on par with the original.
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Tks mirosol :)
ReplyDeleteBuilt this pedal and am very impressed. It gives single coil pickups a full rich tone. Can also be used with Humbuckers. Thanks for the Post.
ReplyDeleteToday, #4
ReplyDeletehttp://mirosol.kapsi.fi/varasto/boxes/SDBoost.jpg
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Mine works fine, but it pops when I turn on the pedal: (
ReplyDeleteI use this wiring:
ReplyDeletehttp://www.beavisaudio.com/techpages/StompboxWiring/
There is no pulldown resistor on the schematic, nor on the layout. Place 1M resistor from input to ground. That should take care of it.
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Ok mirosol I will try this :)
ReplyDeleteCool, Fixed :)
ReplyDeletewhen I can, I put here some pictures of him :)
Here are the photos of my pickup booster :)
ReplyDeletehttps://dl.dropbox.com/u/51252869/Pedais/DSC_5180_800.jpg
https://dl.dropbox.com/u/51252869/Pedais/DSC_5172_800.jpg
Regards
Continues with "pops": (
ReplyDeleteYesterday was no longer, I do not understand
Was it to be longer connected, and capacitors were accumulated with more power?
not tested yesterday almost no time: (
It could also be the outpout. So pulldown resistor there could also help... Can't remember that mine was popping though.
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Already tested on input and output. With LED and without LED, continues forever.
ReplyDeleteWhat happens is, when I turn on, it does "pop", then stay on a while, if I turn off and on makes no noise. If you let off a little and turn on it back to make pop.
Need help
Fixed finally
ReplyDeleteI put a resistance in input and output the 200k, but as a rule they told me, I should be put in 100k output, and an resistance the 1M in input.
For now good with 200k :)
Good to hear you solved it! Usually 1M works in that king of situations, but for future builders, it's good to know that this one may need ~200K.
DeleteThanks Daniel!
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Great!! I have the original SD Pick Booster but want make one this and compare...can any one post sounds??? :-) Nice pics of Daniel Ludgero Pickup Booster!
ReplyDeleteI dont like the original with SMD componets....is ugly
ReplyDeleteis possible by placing the second switch position, with more bass?
ReplyDeleteTry swapping the 1n and 3n3 for bigger values.
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ReplyDeleteI'm using an original SD pickup booster in the effects loop of an Orange OR15. The minimum volume boost is much too much. Is there a simple way of decreasing the amount of boost?
ReplyDeleteThanks,
Andy
Probably a silly question, but am I right in thinking that the shunted cap at the input trick (=resonance control here) would work at the input of any booster/pedal in general to mellow down the tone at the source?
ReplyDeleteSD has a new version of their pickup booster. You might wanna check it out.
ReplyDeleteHello guys, i only have 2N2907 and 2N3906, do you think will work fine with any of these?
ReplyDeleteEither of those will work fine for Q3. For Q1 and Q2 you'll need NPN transistors.
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Thanks a lo Miro!
DeleteFirst, thanks for all the great posts.
ReplyDeleteWith all the replies and folks claiming to have built the circuit, I have to
suspect that my question will display my electronic ignorance, but here goes...
In the diagram, Q1 and Q2 are NPN transistors with a pinout of CBE, but then
pin 3 (the emitter) of Q2 is tied directly to +9V. This would only make sense
to me if pin 3 is the collector or the view is from the bottom. What am I
missing?
Is there a schematic available? Is there any interest/activity on the newer
0dB model?
TIA
Q3 is PNP so the usual convention is reversed, emitter goes to the higher voltage, collector lower.
DeleteYou can find the schematic here.
My question is/was about Q1 and Q2. Thanks also for the link, though.
ReplyDeleteYou are missing the Q3 on that equation. If the Q1 and Q2 were wired as cascading amplifiers, then your question would be valid. But they are not.
DeleteI've built a couple of these over the years and the layout plus the schematic are working as it should.
+m
Q1 and Q2 look like a differential pair to me
Delete"In the diagram, Q1 and Q2 are NPN transistors with a pinout of CBE, but then
Deletepin 3 (the emitter) of Q2 is tied directly to +9V."
Q3 emitter is tied to 9v, Q2 collector is tied to 9v and Q1 collector connects to 9v via 30k resistor.
This is getting Kafkaesque! In the diagram, the topmost pin of Q2 should be pin 3, yes? If the transistor is CBE, then the topmost pin is the Emitter, yes? It is tied to +9V, yes?
ReplyDeleteI know I am missing something. Everybody can't be wrong.
Thanks for all your time.
The top pin of Q2 and Q1 are the collectors. They're labelled C. Q2 collector is tied directly to 9V, and the emitters of both Q1 and Q2 are tied to ground through a 33K resistor
ReplyDeleteQ3 is the only one that's "backwards" (because it's PNP) as Mark and Miro explained
Travis,
ReplyDeleteIf the top pin of Q1 and Q2 are collectors, then doesn't that make the pinout EBC and not CBE?
I couldn't find a BC107 pinout for the TO92 package. However, the BC548, which is a plastic TO-92 equivalent of the BC107 according to Wiki, is a CBE.
I am starting to think that Q1 and Q2 are reversed. Of course, it's hard to believe that this didn't get caught when folks built the circuit, so I still might be bonkers.
Please take a look and let me know...
It seems like you're caught up on the orientation of the transistor symbol, when the only thing you need to pay attention to are the C, B, and E locations
ReplyDeleteWhether the transistor is CBE or EBC is irrelevant. Just insert it so that the collector goes to C, base to B, and emitter to E
Aha! You are looking at the schematic and I am looking at the diagram of the Vero board.
ReplyDeleteHehehe!
So you're suggesting the vero and schematic do not match each other?
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Miro,
DeleteYes, unless I'm crazy. Q1 and Q2 need to be turned around to face the same way as Q3. Then, their pin 1 (collector) will then be connected to +V.
I'm pretty sure at this point, but you need to sanity check me, of course.
Maybe the best course would be to swap out the TO92 packages for TO18s if possible. I am guessing that when everyone built the boards they used TO18 parts and thus didn't detect the orientation.
Of course, I could still be wrong. You be the judge.
Many thanks again for all your work. It is impressive and much appreciated.
I'm sorry to tell you but you are wrong and you have been all the way. Just look at the layout and look at the schematic. On both, Q1 has 30K between its collector and supply, right? On both, Q1 has 30K between its emitter and ground, right? On both, Q2 has its collector to supply, right? Emitter of Q2 is connected to Q1 emitter and thus, via 30K to ground, right? Emitter of Q3 is connected to supply, right? Layout has to match the schematic for it to work. If you are convinced that the layout is in error, then the Analogguru's schematic has to be in error.
DeleteYou should probably stop using pin numbers from transistor leads as they vary between 2N and BC models and are not standard. Just use the names - base, collector and emitter.
And no. I've built this circuit twice with TO92 transistors...
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Yes you're getting too hung up on the transistor graphic. The notes say that Q1 and Q2 are BC107's which are round metal can transistors and don't match the layout, but it suggests trying BC550's (and BC560s for Q3) which would be inserted exactly as shown in the layout. Q1 and Q2 with flat side to the left makes the pin out C, B and E from top to bottom as shown in the layout. Q3 with the flat side to the right makes it E, B and C from top to bottom, again exactly as shown in the layout.
DeleteThe orientation of the transistors is always going to depend on what you use. If you used 2N5088's instead then the transistors would all need rotating 180 degrees to those shown in the layout. But again as Miro and Travis have both said, it's the pin out that is important so make sure that matches the order shown in the layout no matter what you use.
Miro,
ReplyDeleteI bow to your expertise although I still can't see it. If you have checked it and are satisfied, then we're good.
If I may ask, what software are you using for your Vero layouts? I've looked around the site, but didn't see any posts.
TIA
It is rather small circuit, so building it isn't too big of a task. That way you could see it works for yourself.
DeleteApp:
https://code.google.com/p/diy-layout-creator/
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Thanks to everyone. I must be too dense to grasp the concept of CBE. Everyone can't be wrong, right? It must be me.
ReplyDeleteI had my say and I think that everything productive and positive has been explored.
I apologize if I have injected static into the forum. It wasn't my intent.
Everything is good.
Thanks once again, Mirosol. I'll give it a try.
ReplyDeleteehy guys...50k pot should works good?
ReplyDeletethanks.
i would solder a 50k resistor between lugs 1 and 3 to make it kinda 25k pot. 50k pot might distort after 12 o'clock settings.
DeleteThis comment has been removed by the author.
ReplyDeleteOk great!
ReplyDeleteBut now...
i have BC177, BC177A and BC177B
and also BC107A and BC107B
What do you suggest?
I know that i should socket and try...but i'm putting the circuit onboard my stratocaster, under the pickguard, and trying it's too hard..
:(
Hey! somebody here said that minimal 6db of boost is to much for him. But what if we just change boost pot on link and add master volume pot after output of circuit? if my idea is correct it will be same boost pedal but with opportunity of any wanted level of volume. what do you think about that idea?
ReplyDeleteso say I wanted to put this into a guitar. I'm using a B500k push pull. I used a 51k metal film resistor and soldered the outer lugs giving me the resistance of approximately 45k to "solve" the decibel issue.
ReplyDeleteHowever I'm having a hard time wrapping my head around the last little bit. How would I be able to switch the battery on and off. It's an onboard preamp I'm trying for. I tried the "true bypass" method below which I know it makes no sense considering this pedal is better as a buffer but as we can tell I'm really new to this.
the top method of 'true bypass" i went for
http://www.muzique.com/schem/dpdt.gif
so I get my guitars sound properly , all the connections are fine. That is when I pull up on the push pull.
when i have it down I get no sound at all.
should I just have the switch for the battery on and off? or what would you guys do.
This comment has been removed by the author.
DeleteI wired the B500k pot backwards to get my way around the "C" pot taper type , the decibel issue as in going from 3 to 25db instead of 6db to 25db.
Deletebut would i do with a push pull to get it to work.
hello, here is my build of this pedal
ReplyDeletehttp://hpics.li/c003e5a
http://hpics.li/730a8e8
this is my first build, but I am quite happy with he result.
I use it between my guitar and my tube amp (Kustom defender) and it really takes it to a new level of distortion that I was waiting for.
The case is a cooking device from French army. The knob comes from an old tube radio receiver, and the light comes from a new old stock I got.
very simple circuit
ReplyDeleteHello I built it but there is a little boost maybe a fault and I am not an expert.q1 9-4-4v q2 9-4-4v q3 9-9-4v is this right?
ReplyDelete